In an arbitrary ring, I have that $ab = 1$. I'm pretty certain that $a(-b) = -1$, but I'm not sure how to prove that the $-$ sign commutes. Is this necessarily true? Could someone point me in the direction of a proof or counterexample?
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2It's a general fact that $a(-b)=(-a)b=-(ab)$. – egreg Jan 16 '17 at 23:03
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Hint: You want to prove $a(-b)=-(ab)$. That is, you want to prove $a(-b)$ is the additive inverse of $ab$. So try adding $a(-b)$ and $ab$ and see if you can show the sum is $0$.
Eric Wofsey
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@Alex $,a(-b),$ and $,(-a)b,$ are both inverses of $,ab,$ so are equal by uniqueness of inverses. which yields a proof of the Law of Signs. – Bill Dubuque Jan 16 '17 at 23:28